Which of the following complex ions is diamagnetic ?

Options

A. `[FeF_(6)]^(3-)`
B. `[CoF_(6)]^(3-)`
C. `[Co(C_(2)O_(4))_(3)]^(3-)`
D. `[Fe(CN)_(6)]^(4-)`

Explanation

To determine which of the following complex ions is diamagnetic, we need to analyze the electronic configurations and the number of unpaired electrons in each complex ion. A complex is considered diamagnetic if all of its electrons are paired.

### Step-by-Step Solution:

1. **Identify the Complex Ions**:
We have four complex ions to analyze. Let's denote them as:
- (i) [FeF6]³⁻
- (ii) [CoF6]³⁻
- (iii) [Co(C2O4)]³⁻
- (iv) [Fe(CN)6]²⁻

2. **Determine the Oxidation State**:
- For [FeF6]³⁻: The oxidation state of Fe is +3.
- For [CoF6]³⁻: The oxidation state of Co is +3.
- For [Co(C2O4)]³⁻: The oxidation state of Co is +3.
- For [Fe(CN)6]²⁻: The oxidation state of Fe is +2.

3. **Electronic Configuration**:
- For Fe (atomic number 26):
- Ground state: [Ar] 4s² 3d⁶
- For Fe³⁺: Remove 3 electrons → [Ar] 3d⁵
- For Fe²⁺: Remove 2 electrons → [Ar] 3d⁶
- For Co (atomic number 27):
- Ground state: [Ar] 4s² 3d⁷
- For Co³⁺: Remove 3 electrons → [Ar] 3d⁶

4. **Analyze the Ligands**:
- Fluoride (F⁻) is a weak field ligand and does not cause pairing of electrons.
- Oxalate (C2O4²⁻) is also a weak field ligand and does not cause pairing of electrons.
- Cyanide (CN⁻) is a strong field ligand and causes pairing of electrons.

5. **Count Unpaired Electrons**:
- For [FeF6]³⁻: 3d⁵ → 5 unpaired electrons (Paramagnetic)
- For [CoF6]³⁻: 3d⁶ → 4 unpaired electrons (Paramagnetic)
- For [Co(C2O4)]³⁻: 3d⁶ → 4 unpaired electrons (Paramagnetic)
- For [Fe(CN)6]²⁻: 3d⁶ → 0 unpaired electrons (Diamagnetic)

6. **Conclusion**:
The only complex ion that is diamagnetic (with all electrons paired) is [Fe(CN)6]²⁻.

### Final Answer:
The complex ion that is diamagnetic is **[Fe(CN)6]²⁻**.